reference of a variable passed to function / method

>>When &$variable is passed to a function, that means that any operations on $variable are affecting the original variable.

does that mean that both function and $variable are altered?

<?php
// A) Parameters passed by value
function swap($a, $b)
{
    $temp = $a;
    $a = $b;
    $b = $temp;
}
// B) Parameters passed by reference
function swap2(&$a, &$b)
{
    $temp = $a;
    $a = $b;
    $b = $temp;
}

$x = 5;
$y = 6;

echo'// 1. Swap using value<br>';
swap($x, $y);
echo "X: $x<br/>";
echo "Y: $y<br/>";

echo'// 2. Swap using call-time pass-by-reference (Deprecated as of PHP 5.3)<br>';
swap(&$x, &$y);
echo "X: $x<br/>";
echo "Y: $y<br/>";

echo'// 3. Swap using reference parameters<br>';
swap2($x, $y);
echo "X: $x<br/>";
echo "Y: $y<br/>";

echo'//4. Swap using reference of reference parameters<br>';
swap2(&$x, &$y);
echo "X: $x<br/>";
echo "Y: $y<br/>";
?>

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output:
// 1. Swap using value
X: 5
Y: 6
// 2. Swap using call-time pass-by-reference (Deprecated as of PHP 5.3)
X: 6
Y: 5
// 3. Swap using reference parameters
X: 5
Y: 6
//4. Swap using reference of reference parameters
X: 6
Y: 5


what does part 4 do?
echo'//4. Swap using reference of reference parameters<br>';
swap2(&$x, &$y);
echo "X: $x<br/>";
echo "Y: $y<br/>";


>>By default, function arguments are passed by value (so that if the value of the argument within the function is changed, it does not get changed outside of the function). To allow a function to modify its arguments, they must be passed by reference.

So which one is inside, outside, (inside and outside)?

so &$ is actual variable and $ is just a copy which does not affect original variable


thanks

&$x is the pointer to the data that is contained in $x.  A pointer is the information that is in the symbol table.  The data is the content of the variable that is pointed to by the symbol table.